The Diophantine equation,
$$x_1^4+x_2^4+x_3^4+x_4^4 = z^4\tag1$$
can be solved with $x_1 = 0$ (Elkies), or as the title shows, with $x_1 = 2$ (Wroblewski). See link here.
Q: Can $(1)$ be solved in the integers with $x_1 = 1$,
$$1+x_2^4+x_3^4+x_4^4 = z^4\tag2$$
or is there a congruential constraint that prevents this?
P.S. Anyone has the computing power to extend Wroblewski's table? He created that back in 2009 (with slower computers). One helpful fact is that all primitive solutions have the form,
$$x_0^4+5^4(x_1^4+x_2^4+x_3^4) = y_1^4$$
or three terms are always multiples of $5$.